The "filmstrip" represents five molecular scenes of a gaseous mixture as it reaches equilibrium over time:

X is purple and Y is orange: ${\mathbf{X}}_{\mathbf{2}}\left(g\right)\mathbf{+}{\mathbf{Y}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}\mathbf{XY}\left(g\right)$.

(a) Write the reaction quotient, Q, for this reaction.

(b) If each particle represents$\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{mol}$, find Q for each scene.

(c) If K>1, is time progressing to the right or to the left? Explain.

(d) Calculate K at this temperature.

(e) If$\mathbf{∆}{\mathbf{H}}_{\mathbf{rxn}}^{\mathbf{o}}$, which scene, if any, best represents the mixture at a higher temperature? Explain.

(f) Which scene, if any, best represents the mixture at a higher pressure (lower volume)? Explain.

Reactions: A balanced chemical reaction equation shows the mole relationships of reactants and products, while a chemical reaction equation gives the reactants and products. The amount of energy involved in the reaction is frequently stated. Reaction stoichiometry is the study of the quantitative aspects of chemical reactions.

(a)

Considering the given information:

$$\begin{gathered} {\mathrm{X}}_2\left(\mathrm{g}\right)+{\mathrm{Y}}_2\left(\mathrm{g}\right)\to 2\mathrm{XY}\left(\mathrm{g}\right) \\ \mathrm{Q}=? \end{gathered}$$

For the given reaction, the reaction quotient is:

$\mathrm{Q}=\frac{{\left[\mathrm{XY}\right]}^2}{\left[{\mathrm{X}}_2\right]\left[{\mathrm{Y}}_2\right]}$

(b)

Considering the given information:

X - purple colour spheres

Y - orange colour spheres

Scene A:

$$\begin{gathered} \left[{\mathrm{X}}_2\right]=0.4\mathrm{mol} \\ \left[{\mathrm{Y}}_2\right]=0.4\mathrm{mol} \\ \left[\mathrm{XY}\right]=0.0\mathrm{mol} \end{gathered}$$

Scene A's reaction quotient is:

$$\begin{gathered} \mathrm{Q}=\frac{\left[\mathrm{XY}\right]}{\left[{\mathrm{X}}_2\right]\left[{\mathrm{Y}}_2\right]} \\ \mathrm{Q}=\frac{0}{0.4\times 0.4} \\ \mathrm{Q}=0 \end{gathered}$$

Scene B:

$$\begin{gathered} \left[{\mathrm{X}}_2\right]=0.2\mathrm{mol} \\ \left[{\mathrm{Y}}_2\right]=0.2\mathrm{mol} \\ \left[\mathrm{XY}\right]=0.4\mathrm{mol} \end{gathered}$$

Scene B's reaction quotient is:

$$\begin{gathered} \mathrm{Q}=\frac{{\left[\mathrm{XY}\right]}^2}{\left[{\mathrm{X}}_2\right]\left[{\mathrm{Y}}_2\right]} \\ \mathrm{Q}=\frac{{\left[0.4\right]}^2}{0.2\times 0.2} \\ \mathrm{Q}=4 \end{gathered}$$

Scene C:

$$\begin{gathered} \left[{\mathrm{X}}_2\right]=0.6\mathrm{mol} \\ \left[{\mathrm{Y}}_2\right]=0.1\mathrm{mol} \\ \left[\mathrm{XY}\right]=0.1\mathrm{mol} \end{gathered}$$

Scene C's reaction quotient is:

$$\begin{gathered} \mathrm{Q}=\frac{{\left[\mathrm{XY}\right]}^2}{\left[{\mathrm{X}}_2\right]\left[{\mathrm{Y}}_2\right]} \\ \mathrm{Q}=\frac{{\left(0.6\right)}^2}{0.1\times 0.1} \\ \mathrm{Q}=36 \end{gathered}$$

Because these scenes have the same number of moles, the reaction quotient for scenes D and E is equal to the reaction quotient for scene C.

(c)

Considering the given information:

Q=0 in the scene A

If $\mathrm{K}>1$, time should be advanced to the right in order to achieve equilibrium.

(d)

Considering the given information:

Q=K at equilibrium

The mixture reaches equilibrium at scene C because the moles of reactants and products do not change from scene C.

Therefore, at this temperature, $\mathrm{K}=36$.

(e)

Considering the given information:

When$∆{\mathrm{H}}_{\mathrm{rxn}}^{\mathrm{o}}$ the value is negative, the products have less energy than reactants.

The forward reaction - exothermic reaction

The reverse reaction - endothermic reaction

The products absorb heat and convert into reactants at high temperatures.

So in this case, scene A is the best representation.

(f)

When the number of moles of reactants and products is equal, pressure has no effect on the reaction, so no scene best represents the mixture at a higher pressure.